B. Dynamics · Dynamics: Forces and Motion in a PlaneRelative Velocity: Rivers, Winds and Moving Observers
B2.2 B2.1 A1.12About 70 minutes
Learning goals- I can use subscript notation, v(AC) = v(AB) + v(BC), to combine relative velocities.
- I can find the time, drift and heading for a boat crossing a river with a current.
- I can find the heading and ground speed an aircraft needs to follow a chosen track in a crosswind.
- I can find the velocity of one moving object as seen from another, v(AB) = v(A) − v(B).
Words to know
frame of reference
A coordinate system (and the observer attached to it) from which positions and velocities are measured, for example the ground, the water, or a moving car.
relative velocity
The velocity of an object as measured by a particular observer or frame of reference.
heading
The direction an aircraft or boat is pointed. Because of wind or current, it is not always the direction it actually travels.
airspeed
The speed of an aircraft relative to the air around it.
ground velocity
The velocity of an object relative to the ground (Earth's surface).
Why is the flight home faster?
Across Canada, eastbound flights are usually quicker than westbound flights on the same route. The plane is not faster; the air is moving. High-altitude winds (the jet stream) generally blow from west to east, so an eastbound plane is carried along, while a westbound plane fights a headwind.
Velocity always depends on who is measuring it. A passenger walking down the aisle at 1.5 m/s is walking at 1.5 m/s relative to the plane, but at hundreds of kilometres per hour relative to the ground. This lesson gives you a reliable way to keep the observers straight.
Subscript notation
Write v(AB) for "the velocity of A relative to B". Then velocities chain together like links:
v(AC) = v(AB) + v(BC)
The inside letters match (B and B), and the outside letters give the answer (A and C). Two more useful facts:
- v(BA) = −v(AB). If you see me moving east at 3 m/s, I see you moving west at 3 m/s.
- For two objects both measured relative to the ground (G): v(AB) = v(AG) − v(BG).
The addition is vector addition. When the velocities are not along one line, draw them tip to tail and solve with components, or with Pythagoras and trigonometry when there is a right angle.
Remember from Grade 11
In SPH3U you added a canoe's velocity and a river's current tip to tail, and learned that the crossing time depends only on the velocity component straight across the river. Today adds subscripts, aiming a craft to reach a chosen point, and the velocity of one moving object seen from another.
Worked exampleExample 1: Crossing the river two ways
A river 120 m wide flows east at 1.5 m/s. A canoeist can paddle at 2.5 m/s relative to the water. (a) She points the canoe straight across [N]. Find her ground velocity, crossing time and how far downstream she lands. (b) Instead she wants to land directly opposite her starting point. Which way must she point the canoe, and how long does the crossing take?
- Let C = canoe, W = water, G = ground. v(CG) = v(CW) + v(WG), with v(WG) = 1.5 m/s [E].
- (a) v(CW) = 2.5 m/s [N]. These are perpendicular: |v(CG)| = √(2.5² + 1.5²) = 2.92 m/s; θ = tan⁻¹(1.5 ÷ 2.5) = 31° east of north.
- Only the northward component (2.5 m/s) carries her across: Δt = 120 ÷ 2.5 = 48 s. Drift = 1.5 × 48 = 72 m downstream.
- (b) Now v(CG) must point due north, so the east part of v(CW) must cancel the current: 2.5 sin θ = 1.5, so sin θ = 0.60 and θ = 37° west of north.
- Ground speed = √(2.5² − 1.5²) = 2.0 m/s [N]. Δt = 120 ÷ 2.0 = 60 s.
(a) 2.9 m/s [N 31° E], 48 s, landing 72 m downstream. (b) Point the canoe [N 37° W]; the crossing takes 60 s. Aiming straight across is the fastest crossing; aiming upstream is the shortest path.
In (a) the canoe is pointed straight across and the current carries it downstream. In (b) it is pointed upstream at 37° so that the current's push is cancelled and the ground velocity is due north. In both, the dashed arrow is the current, 1.5 m/s [E].Worked exampleExample 2: Flying a straight track in a crosswind
A small plane has an airspeed of 220 km/h. The wind blows at 60 km/h [E]. The pilot wants to travel due north. Find the heading and the ground speed.
- v(PG) = v(PA) + v(AG). We want v(PG) to point [N]. v(AG) = 60 km/h [E].
- The plane must point partly west so that its west component cancels the wind: 220 sin θ = 60, so θ = sin⁻¹(60 ÷ 220) = 15.8° west of north.
- Ground speed = √(220² − 60²) = √(48 400 − 3600) = 212 km/h.
Heading [N 16° W], ground speed about 212 km/h [N].
Worked exampleExample 3: Wind at an angle (components)
A plane with an airspeed of 200 km/h must fly a track due east. The wind is 50 km/h [S 30° E]. Find the heading and the ground speed.
- Wind components (east +x, north +y): x = 50 sin 30° = 25.0 km/h; y = −50 cos 30° = −43.3 km/h.
- For a track due east, the ground velocity has y = 0. So the plane's own y component must be +43.3 km/h.
- Plane's x component: √(200² − 43.3²) = 195.3 km/h. Heading: θ = tan⁻¹(43.3 ÷ 195.3) = 12.5° north of east.
- Ground velocity: x = 195.3 + 25.0 = 220.3 km/h; y = 0.
Heading [E 12.5° N]; ground speed about 220 km/h [E]. The wind both pushes the plane off course and gives it a small tailwind.
One moving object seen from another
Two snowmobiles leave a crossroads: A goes 15 m/s [N] and B goes 20 m/s [E]. What does the rider on B see A doing?
v(AB) = v(AG) − v(BG) = 15 [N] + 20 [W] (the opposite of B's velocity).
|v(AB)| = √(15² + 20²) = 25 m/s, at tan⁻¹(20 ÷ 15) = 53° west of north: [N 53° W]. To B's rider, A seems to be moving away diagonally backward.
The same idea explains why rain that falls straight down hits your car's windshield at a slant. In the car's frame, the rain has an extra horizontal velocity opposite to the car's motion.
Just for funA plane flying at 900 km/h into a 150 km/h headwind and a plane flying at 600 km/h with a 150 km/h tailwind travel at the same ground speed. To the people on the ground, they are perfectly matched; to the pilots, one is working much harder than the other.
Hands-on activityRelative motion on a moving walkway (or a hallway)
You need: a long hallway or driveway, a measuring tape, masking tape or chalk, a stopwatch (phone), a toy car or a ball, and a friend or family member.
- Mark a 5.0 m track on the floor. Have your partner walk the length of it at a steady pace while carrying a tray or a large book. Time it and find their speed relative to the floor.
- Roll a ball or toy car across the tray (sideways to the direction of walking) while your partner walks. From your partner's point of view, the ball rolls straight across the tray. Watch it from the side: what path does it follow relative to the floor?
- Sketch the vector triangle: v(ball, tray) + v(tray, floor) = v(ball, floor).
- Estimate the ball's speed across the tray (tray width ÷ time to cross). Calculate the ball's speed and direction relative to the floor, and compare with what you saw.
Walk slowly and clear the path of tripping hazards.
Your turn
1A passenger walks at 1.2 m/s toward the back of a train moving 25 m/s [E]. Find her velocity relative to the ground.
v(PG) = v(PT) + v(TG) = 1.2 [W] + 25 [E] = 23.8 m/s [E].
2A boat that moves at 4.0 m/s relative to the water heads straight across an 80 m wide river whose current is 3.0 m/s. Find its ground speed, crossing time and downstream drift.
Ground speed = √(4.0² + 3.0²) = 5.0 m/s. Time = 80 ÷ 4.0 = 20 s. Drift = 3.0 × 20 = 60 m.
3The same boat wants to land directly opposite. At what angle upstream from straight across must it head, and how long does the crossing take?
sin θ = 3.0 ÷ 4.0 = 0.75, so θ = 49° upstream. Ground speed = √(4.0² − 3.0²) = 2.65 m/s. Time = 80 ÷ 2.65 = 30 s.
4A plane with an airspeed of 300 km/h heads [N]. The wind is 40 km/h [E]. Find the ground velocity.
√(300² + 40²) = 303 km/h; θ = tan⁻¹(40 ÷ 300) = 7.6°: [N 7.6° E].
5If v(AB) = 5.0 m/s [S], what is v(BA)?
5.0 m/s [N] (equal and opposite).
6Car A moves 25 m/s [E] and car B moves 20 m/s [W] in the other lane. Find the velocity of A relative to B.
v(AB) = v(AG) − v(BG) = 25 [E] − 20 [W] = 25 [E] + 20 [E] = 45 m/s [E].
More practice: warm-up
1What does v(AB) mean in words?
The velocity of A relative to B (as B would measure it).
2Complete: v(AC) = v(A?) + v(?C).
3A person sits still on a bus moving 15 m/s [E]. What is the person's velocity relative to the bus?
4A cyclist rides at 8 m/s [E] into a 3 m/s wind blowing [W]. What is the air's velocity relative to the cyclist?
v(air, cyclist) = 3 [W] − 8 [E] = 11 m/s [W]: she feels an 11 m/s headwind.
5For a boat crossing a river, which velocity component sets the crossing time?
The component straight across the river (perpendicular to the current).
More practice: on your own
1A pilot wants a track due east. The airspeed is 250 km/h and the wind is 50 km/h [S]. Find the heading and ground speed.
The plane must point partly north to cancel the wind: sin θ = 50 ÷ 250 = 0.20, θ = 11.5°. Heading [E 12° N] (11.5°). Ground speed = √(250² − 50²) = 245 km/h.
2Rain falls straight down at 6.0 m/s. A car drives at 15 m/s. At what angle from the vertical do the rain streaks appear on the side windows?
In the car's frame the rain has 15 m/s horizontal (backward) and 6.0 m/s down. θ = tan⁻¹(15 ÷ 6.0) = 68° from the vertical, slanting toward the back.
3A swimmer can swim 1.2 m/s in still water. A river 50 m wide flows at 0.80 m/s. She wants to cross in the least time. Which way should she head, how long does it take, and how far downstream does she land?
Head straight across: the whole 1.2 m/s works against the width. Δt = 50 ÷ 1.2 = 42 s (41.7). Drift = 0.80 × 41.7 = 33 m downstream.
4Ship A sails 10 km/h [N] and ship B sails 10 km/h [E]. Find the velocity of B relative to A.
v(BA) = v(BG) − v(AG) = 10 [E] + 10 [S]. Magnitude = √(10² + 10²) = 14 km/h, direction [S 45° E].
5A plane heads [N 30° E] with an airspeed of 400 km/h. The wind is 80 km/h [W]. Find the ground velocity.
Plane: x = 400 sin 30° = 200; y = 400 cos 30° = 346.4. Add the wind: x = 200 − 80 = 120; y = 346.4. |v| = √(120² + 346.4²) = 367 km/h; θ = tan⁻¹(120 ÷ 346.4) = 19°: [N 19° E].
Challenge
1A plane with an airspeed of 180 km/h must fly 500 km due north. The wind blows at 40 km/h [E 30° N]. Find the heading and the time for the trip.
Wind: x = 40 cos 30° = 34.6; y = 40 sin 30° = 20.0. The plane's x must be −34.6: sin θ = 34.6 ÷ 180, θ = 11.1° west of north. Plane's y = √(180² − 34.6²) = 176.6. Ground speed = 176.6 + 20.0 = 196.6 km/h. Heading [N 11° W]; time = 500 ÷ 196.6 = 2.5 h (2.54).
2A plane with an airspeed of 200 km/h flies 600 km straight into a 50 km/h headwind and then 600 km back with the wind behind it. Compare the round-trip time with the time on a calm day, and explain the result.
Out: 600 ÷ 150 = 4.0 h. Back: 600 ÷ 250 = 2.4 h. Total 6.4 h, versus 1200 ÷ 200 = 6.0 h with no wind. The plane spends more time flying slowly against the wind than it spends flying fast with it, so the wind's help never fully makes up for its hindrance.
3Rain falls vertically. A driver moving at 18 m/s sees the streaks make a 60° angle with the vertical. How fast is the rain falling?
tan 60° = (horizontal speed) ÷ (vertical speed) = 18 ÷ v. v = 18 ÷ tan 60° = 18 ÷ 1.732 = 10 m/s (10.4).
Quiz
1. Using subscripts, v(PG) = v(PA) + ?
2. A canoe aimed straight across a river with a current:
3. A plane must fly due north while the wind blows east. The pilot should head:
4. Car A moves 20 m/s [E] and car B moves 15 m/s [E]. The velocity of A relative to B is:
5. A boat moving 3.0 m/s relative to the water heads straight across a river with a 4.0 m/s current. Its ground speed is:
6. Rain that falls straight down appears to hit a moving car's side window at a slant because:
Big idea: Velocities depend on the observer's frame of reference, and relative velocities combine by vector addition (B2.2). This lesson extends SPH3U river problems to aiming upstream, aircraft in crosswinds and the velocity of one moving object as seen from another. It sets up the lesson Inertial and Non-inertial Frames of Reference later in this unit.
How to help: Ask your teen to name each velocity aloud with its subscripts ("canoe relative to water", "water relative to ground") before drawing. Most errors come from adding the wrong pair.
Common difficulties:
- Confusing the heading (where the craft points) with the track (where it goes).
- Using Pythagoras when the triangle has no right angle (use components instead).
- Adding instead of subtracting for v(AB) = v(AG) − v(BG).
- Thinking a current makes the crossing slower when the boat aims straight across (it does not).